Which one of the following will have the largest number of atoms?
- 1 g Au(s)
- 1 g Na(s)
- 1 g Li (s)
- 1 g of $C{{l}_{2}}$ (g)
- 1 g Au = 1/197 mote atoms of
Au = 1/197 x 6.022 x ${{10}^{23}}$ atoms of Au.
-
1 g Na = 1/23 mole atoms of Na = 1/23 x 6.022 x ${{10}^{23}}$ atoms of Na.
-
1 fg Li = 1/7 mole atoms of
Li = 1/7 x 6.022 x ${{10}^{23}}$ molecules of Li. -
1 g $C{{l}_{2}}$ = 1/71 mole molecules of
$C{{l}_{2}}$ = 2/71 x 6.022 x ${{10}^{23}}$ atoms Cl.